Question

The speed of light in media ‘A’ and ‘B’ are 2.0 × 10¹⁰ cm/s and 1.5 × 10¹⁰ cm/s respectively. A ray of light enters from the medium B to A at an incident angle ‘θ’. If the ray suffers total internal reflection, then

• A. \(θ = sin^{-1} (\frac{3}{4})\)
• B. \(θ > sin^{-1} (\frac{2}{3})\)
• C. \(θ < sin^{-1} (\frac{3}{4})\)
• D. \(θ > sin^{-1} (\frac{3}{4})\)

Answer 1

The Correct Option is D

 To solve the problem, we must understand the concept of total internal reflection and the physics behind it.

When a light ray attempts to move from a medium of higher refractive index to one of lower refractive index (from medium B to A in this case), total internal reflection occurs if the angle of incidence is greater than a certain critical angle.

Given:

• Speed of light in medium A: \( v_A = 2.0 \times 10^{10} \) cm/s
• Speed of light in medium B: \( v_B = 1.5 \times 10^{10} \) cm/s

We can calculate the refractive indices of the media using the formula:

\(n = \frac{c}{v}\)

where \( n \) is the refractive index, \( c \) is the speed of light in a vacuum (\( 3.0 \times 10^{10} \) cm/s), and \( v \) is the speed of light in the medium.

Refractive index of medium A, \( n_A \):

\(n_A = \frac{3.0 \times 10^{10}}{2.0 \times 10^{10}} = 1.5\)

Refractive index of medium B, \( n_B \):

\(n_B = \frac{3.0 \times 10^{10}}{1.5 \times 10^{10}} = 2.0\)

Total internal reflection occurs when the angle of incidence is greater than the critical angle, which is determined by:

\(\sin \theta_c = \frac{n_A}{n_B} = \frac{1.5}{2} = 0.75\)

The critical angle, \( \theta_c \), is therefore:

\(\theta_c = \sin^{-1}(0.75)\)

Total internal reflection occurs if the angle of incidence \( \theta \) is greater than \( \theta_c \). Thus, the condition for total internal reflection is:

\(\theta > \sin^{-1}\left(\frac{3}{4}\right)\)

Hence, the correct option is:

1. \( \theta > \sin^{-1} \left( \frac{3}{4} \right) \)